An amount of of is dissolved in water and the total volume made up to 500 ml. What is the percentage — Ionic Equilibrium Chemistry Question
Question
An amount of $0.16 \text{ g}$ of $\text{N}_2\text{H}_4$ is dissolved in water and the total volume made up to 500 ml. What is the percentage of $\text{N}_2\text{H}_4$ that has reacted with water in this solution? $K_b$ for $\text{N}_2\text{H}_4 = 4.0 \times 10^{-6}$.
💡 Solution & Explanation
Molar mass of $\text{N}_2\text{H}_4 = 32 \text{ g/mol}$. Moles of $\text{N}_2\text{H}_4 = 0.16 \text{ g} / 32 \text{ g/mol} = 0.005 \text{ mol}$. Concentration $C = 0.005 \text{ mol} / 0.5 \text{ L} = 0.01 \text{ M} = 10^{-2} \text{ M}$. Degree of dissociation $\alpha = \sqrt{K_b / C} = \sqrt{4.0 \times 10^{-6} / 10^{-2}} = \sqrt{4.0 \times 10^{-4}} = 2.0 \times 10^{-2}$. Percentage reaction = $\alpha \times 100\% = 2.0 \times 10^{-2} \times 100\% = 2\%$. Therefore, correct answer is C.